Since I'm not that good at (as I like to call it) 'die-hard-mathematics', I've always liked concepts like the golden ratio or the dragon curve, which are easy to understand and explain but are mathematically beautiful at the same time.

$\begingroup$@TheGuywithTheHat That's the reason for the second spike of visits, on August 27. The comment by LTS is from April; back then the traffic was driven by Ycombinator. As a result, this same question made both April 7 and August 27 the two days with most visits to the site.$\endgroup$
– user147263Aug 29 '14 at 18:37

$\begingroup$My first intuitive visualization of this sum was a circle. I wasn't $100\%$ sure that the answer was $1$(long time back, I had never seen an infinite series before, and 1 was just my first immediate thought) :)$\endgroup$
– GuyMar 31 '14 at 17:38

$\begingroup$You don't need to divide the square into such complicated fragments, just use vertical lines (assuming the x coordinate ranges from 0 to 1) at x=1/2, x=1/4, x=1/8 etc. Each time you add 1/2^n to the area, and in the limit you obviously get 1.$\endgroup$
– Maxim UmanskyApr 7 '14 at 4:05

42

$\begingroup$@MaximUmansky: That way you'd just get lines that get closer together and it'll be not as obvious. Here, you see the fractions $\frac{1}{2}$ and $\frac{1}{4}$ in their "standard shape", so what remains must obviously be $\frac{1}{4}$. Then, put the same shapes inside the remaining square which is of the same proportions as the initial one (and it's easily checked that $\frac{1}{2}\cdot\frac{1}{4}=\frac{1}{8}$); you'll get the next smaller square, hidden deeper in the corner. Repeat, and the square will shrink to a tiny dot (not a whole line, which may intuitively seem larger).$\endgroup$
– nobodyApr 9 '14 at 20:09

$\begingroup$@gekkostate You're a bit off. f is the sum of multiple simple waves, all with different frequencies and phase angles. The fourier transform takes a complex wave from a given time period, and gives you the phase angles and frequencies of all of the component waves. f^ is the amplitude of each component wave.$\endgroup$
– JoelApr 7 '14 at 2:15

6

$\begingroup$I'm down-voting this because this animation doesn't include phase information. The phase is the difference between a pulse and cw. If you don't want it, you should use another basis (wavelets).$\endgroup$
– MikhailApr 7 '14 at 3:19

$\begingroup$I saw this animation a long time ago, and it helped me finally understand what the Fourier transformation does, but I still am lost as to how it works...$\endgroup$
– Cole JohnsonApr 11 '14 at 1:21

11

$\begingroup$I don't find this graphical example helpful in explaining the Fourier Transform. If anything, it is going to add to existing confusion people have with the Fourier Series. The sin and cos expression shown is too general to be meaningful, and the graphic shown for the "transform" is really just a visualization of the magnitudes of the Fourier Series coefficients you would have if the partial sum function were presumed to be infinitely periodic. When I calculate the actual Fourier Transform of the segment of function shown, I get something quite different: imgur.com/eQuerJa$\endgroup$
– Adam SmithAug 31 '14 at 18:53

$\begingroup$If you remove the leftmost column, you get a proof that the sum of the first $n$ non-negative even numbers (counting 0) is $n(n-1)$$\endgroup$
– beckoApr 3 '14 at 16:29

27

$\begingroup$I read something about how Galileo noted that the distance a falling body covered over a unit time went as the series of odd numbers and was confused until I realized this.$\endgroup$
– Nick TApr 3 '14 at 17:37

15

$\begingroup$I remember about 5 years back, it was when we were first learning programming. Just simple kids stuff. Loops. We were given a series to print. $1,4,9,16,\dots$ I was a little overworked so I didn't notice the obvious square pattern. I did notice $1,(1+3),(1+3+5),....$ when we were giving in our work, I saw the guy before me had just done $i*i$. I totally went "hoooly shit"$\endgroup$
– GuyApr 6 '14 at 9:40

$\begingroup$I always found the easiest way to think about this was that if you move around a convex shape such that you eventually end up at the start, you must have rotated through exactly 360º. If you walk around any shape, and count anticlockwise rotations as negative, but clockwise rotations as positive, I imagine the angles will still add to 360º (or -360º depending on which direction you take).$\endgroup$
– daviewalesApr 5 '14 at 12:27

16

$\begingroup$I like to view this as a limiting process. Imagine that the picture on the right is the picture on the left zoomed out a great distance.$\endgroup$
– Steven GubkinApr 5 '14 at 14:10

3

$\begingroup$This would be better as an animation (which achieves the zooming effect).$\endgroup$
– Duncan JonesApr 7 '14 at 10:31

1

$\begingroup$Edited. Do have a look. (It may look horrible, since this is the first time I made an animation.)$\endgroup$
– Yiyuan LeeApr 7 '14 at 13:36

While attending an Abstract Algebra course I was given the assignment to write out the multiplication table modulo n. I forgot to do the homework until just before class, but it was so easy to write the program I was able to print the result between classes.

The circular patterns in the tables fascinated me, and compelled me to replace the numbers with colors. The result is a beautiful illustration showing the emergence of primes and symmetry of multiplication.

The colors were chosen to start blue at 1 (cold) and fade to red at n (hot). White is used for zero (frozen), because it communicates the most information about prime factorization.

$\begingroup$This is awesome. Looks very much like a fractal. Is there a name for this fractal?$\endgroup$
– OmnipotentEntityApr 7 '14 at 15:52

$\begingroup$What in specific is it lacking? Correct me if I'm wrong, but if we take the white dots to be the area of the fractal and the red and blue to be the border then it's an object of fractional dimension (either 0d points or 1d lines filling a 2d plane) and self-symmetric.$\endgroup$
– OmnipotentEntityApr 7 '14 at 17:42

27

$\begingroup$@OmnipotentEntity it doesn't qualify as having fractional dimension because the underlying space is discrete. If you use a real space instead - the set of x, y such that x * y = 0 mod n, you get a set of hyperboles (linear in dimension). If you take a limit towards infinite squares, then the count of white squares is equal to the length of the edge whenever the length is prime.$\endgroup$
– John DvorakApr 7 '14 at 18:48

$\begingroup$I just read about qualitative self-similarity which I think describes the relationship between frames in this example, but it isn't continuous in any dimension.$\endgroup$
– Jared DeckardApr 7 '14 at 18:51

This is a neat little proof that the area of a circle is $\pi r^2$, which I was first taught aged about 12 and it has stuck with me ever since. The circle is subdivided into equal pieces, then rearranged. As the number of pieces gets larger, the resulting shape gets closer and closer to a rectangle. It is obvious that the short side of this rectangle has length $r$, and a little thought will show that the two long sides each have a length half the circumference, or $\pi r$, giving an area for the rectangle of $\pi r^2$.

This can also be done physically by taking a paper circle and actually cutting it up and rearranging the pieces. This exercise also offers some introduction to (infinite) sequences.

$\begingroup$This doesn't work for me :( Sadly, I find it non-obvious that the length of the bottom really will converge helpfully to $\pi r$.$\endgroup$
– Nicholas WilsonApr 8 '14 at 16:50

11

$\begingroup$@NicholasWilson By definition, the circumference of a circle is π times the diameter of the circle. Here, we've cut the circle into slices, half flipped one way, half flipped the other. Therefore half the circumference (πd) is on the bottom. Half of the diameter is the radius, r. Does that help?$\endgroup$
– ghoppeApr 8 '14 at 19:29

47

$\begingroup$@ghoppe Nice try! But it's all wiggly. We're lucky that $\sin{\theta} \sim \theta$ as $\theta \rightarrow 0$, so the actual perimeter of the bottom (which is known to be $\pi$) does converge to the width of the rectangle. But - that nice property of $\sin$ is exactly what we're trying to prove!$\endgroup$
– Nicholas WilsonApr 9 '14 at 8:20

12

$\begingroup$@ghoppe What would you visually infer about the perimeter of a Koch snowflake? I've had people assure me "It must be bounded!" Visual inferences are susceptible to error :( You need to invoke some limit arguments to convince me those arcs on the segments do converge. Similarly, inscribing polygons to determine the circumference sounds very dangerous (think what would happen if you tried that with a Koch snowflake) -- but bounding the area above and below by inscribed/exscribed polygons is definitely a sound proof.$\endgroup$
– Nicholas WilsonApr 9 '14 at 17:56

9

$\begingroup$@NicholasWilson coming back to this discussion rather late, but perhaps I can convince you. The length of the bottom is always πr, it doesn't change with iteration so all we have to show is that the shape ends up as a rectangle. This is equivalent to stating that the line between two points on a circle tends to the tangent as the two points get closer together, which I believe is at least one definition of the tangent.$\endgroup$
– Ben RowlandDec 16 '14 at 22:16

Here is a very insightful waterproof demonstration of the Pythagorean theorem. Also there is a video about this.

It can be explained as follows. We seek a definition of distance from any point in $\mathbb{R}^2$ to $\mathbb{R}^2$, a function from $(\mathbb{R}^2)^2$ to $\mathbb{R}$ that satisfies the following properties.

As a result of this, from now on, I will define the distance from any point $(x, y)$ to any point $(z, w)$ as $\sqrt{(z - x)^2 + (w - y)^2}$ and denote it as $d((x, y), (z, w))$. I will also use $d(x, y)$ as shorthand for $d((0, 0), (x, y))$ Since distance satisfies condition 5, for any right angle triangle at all, not just ones whose legs are parallel to the axes, the square of the length of its hypotenuse is equal to the sum of the squares of the lengths of its legs.

This image shows that using that definition of distance, for any right angle triangle whose legs are parallel to the axes and have lengths $x \in \mathbb{R}^+$ and $y \in \mathbb{R}^+$, the area of a square with the hypotenuse as one of its edges is $(x - y)^2 + 2xy = x^2 + y^2 = (d(x, y))^2$. Combining that result with the fact that distance satisfies condition 5, we can show that for any right angle triangle, even with legs non parallel to the axes, the area of a square with its hypotenuse as its edge has an area equal to the sum of the squares of the lengths of its legs.

$\begingroup$This does not actually demonstrate the pythagorean theorem. It is showing that for one single right triangle, the theorem holds. This could be a coincidence. Someone seeing this might think that 3,4,5 is the only pythagorean triple.$\endgroup$
– SparrApr 7 '14 at 16:06

114

$\begingroup$Sparr, you're incorrect. This does indeed demonstrate the pythagorean theorem. What you mean is that it doesn't prove it. All the demonstrations on this page are simply examples, not the infinity of all possibilities. Think before you speak. And all you upvoters, think before you upvote.$\endgroup$
– B TAug 27 '14 at 21:12

12

$\begingroup$@BT, do you consider the fact that $1^3 + 2^3 \neq 18^3$ a demonstration of Fermat's Last Theorem?$\endgroup$
– Daniel McLauryAug 28 '14 at 1:17

21

$\begingroup$@DanielMcLaury Sure it's a demonstration of FLT, but it's not an interesting one. This gif, however, is interesting because it looks cool.$\endgroup$
– Samuel YusimAug 28 '14 at 5:30

8

$\begingroup$@GrumpyParsnip Before you learn why something is true, you might want to do a sanity check to verify that it probably is. This does that, while looking cool. The only thing that I really care about is that it looks cool, though. Mathematics is a very aesthetic thing, so I'm allowed to like something just because it looks nice.$\endgroup$
– Samuel YusimSep 5 '14 at 0:06

$\begingroup$It is interesting to note that Taylor series are relatively very precise near their center, which makes them especially useful in computer science when you need a precise but still fast approximation. For example, processors do not have a cosine function built-in, but will fly through an equation such as x => 1 - (x^2)/(2!) + (x^4)/(4!) - ... and from this equation, you could build a function that accepts a precision parameter so that the call recurses until higher accuracy is achieved. (note: there are many more efficient ways to do this)$\endgroup$
– sleblancApr 3 '14 at 21:55

1

$\begingroup$@sebleblanc It is just amazing that petty much any function can represented with a Taylor series locally$\endgroup$
– joshlkApr 4 '14 at 8:15

3

$\begingroup$@sebleblanc Should have said "[some] processors" instead. x86 processors have had a cosine function for some time now with the FCOS instruction. It doesn't do it with a taylor series as far as I can tell. What I assume it does is use a lookup table followed by some sum/difference formulas.$\endgroup$
– Cole JohnsonApr 4 '14 at 16:05

$\begingroup$You are right that X87 co-processors are omni-present, but there are far more embedded CPUs in the world than there are X86/87 CPUs. Just think of how many devices you interact with that are not strictly computers. ///// From my desk, I can see three computer monitors, one TV, two video game consoles, one oscilloscope, one modern refrigerator, two cordless phones, a wireless mouse, a microcontroller dev board, a Raspberry Pi, my cell phone, my credit card. All of these have a processor that does not have the FSIN instruction... If anything, I should have said [most] instead. ;-)$\endgroup$
– sleblancApr 12 '14 at 14:31

2

$\begingroup$@sebleblanc The x86/87 FPU uses CORDIC which lets you compute trigonometric functions in hardware using only bit shifts and addition.$\endgroup$
– CalmariusAug 27 '14 at 16:57

When I look up "area of a rhombus" on Google images, I find plenty of disappointing images like this one:

which show the formula, but fail to show why the formula works. That's why I really appreciate this image instead:

which, with a little bit of careful thought, illustrates why the product of the diagonals equals twice the area of the rhombus.

EDIT: Some have mentioned in comments that that second diagram is more complicated than it needs to be. Something like this would work as well:

My main objective is to offer students something that encourages them to think about why a formula works, not just what numbers to plug into an equation to get an answer.

As a side note, the following story is not exactly "visually stunning," but it put an indelible imprint on my mind, and affected the way I teach today. A very gifted Jr. High math teacher was teaching us about volume. I suppose just every about school system has a place in the curriculum where students are required learn how to calculate the volume of a pyramid. Sadly, most teachers probably accomplish this by simply writing the formula on the board, and assigning a few plug-and-chug homework problems.

No wonder that, when I ask my college students if they can tell me the formula for the volume of a pyramid, fewer than 5% can.

Instead, building upon lessons from earlier that week, our math teacher began the lesson by saying:

We've learned how to calculate the volume of a prism: we simply multiply the area of the base times the height. That's easy. But what if we don't have a prism? What if we have a pyramid?

At this point, she rummaged through her box of math props, and pulled out a clear plastic cube, and a clear plastic pyramid. She continued by putting the pyramid atop the cube, and then dropping the pyramid, point-side down inside the cube:

She continued:

These have the same base, and they are the same height. How many of these pyramids do you suppose would fit in this cube? Two? Two-and-a-half? Three?

Then she picked one student from the front row, and instructed him to walk them down the hallway:

Go down to the water fountain, and fill this pyramid up with water, and tell us how many it takes to fill up the cube.

The class sat in silence for about a full minute or so, until he walked back in the room. She asked him to give his report.

"Three," he said.

She pressed him, giving him a hard look. "Exactly three?"

"Exactly three," he affirmed.

Then, she looked around the room:

"Who here can tell me the formula I use to get the volume of a pyramid?" she asked.

One girl raised her hand: "One-third the base times the height?"

I've never forgotten that formula, because, instead of having it told to us, we were asked to derive it. Not only have I remembered the formula, I can even tell you the name of boy who went to the water fountain, and the girl who told us all the formula (David and Jill).

Given the upvoted comment, If high school math just used a fraction of the resources here, we'd have way more mathematicians, I hope you don't mind me sharing this story here. Powerful visuals can happen even in the imagination. I never got to see that cube filling up with water, but everything else in the story I vividly remember.

Incidentally, this same teacher introduced us to the concept of pi by asking us to find something circular in our house (“like a plate or a coffee can”), measuring the circumference and the diameter, and dividing the one number by the other. I can still see her studying the data on the chalkboard the next day – all 20 or so numbers just a smidgeon over 3 – marveling how, even though we all probably measured differently-sized circles, the answers were coming out remarkably similar, “as if maybe that ratio is some kind of constant or something...”

$\begingroup$^^1 million upvotes for the stories about maths education. I am writing both of these down for future reference. In fact, I think we should start a website with instructions to help primary teachers teach mathematical concepts in interesting ways.$\endgroup$
– daviewalesApr 5 '14 at 12:34

$\begingroup$Great stories. I do, however, think your first rhombus is fine. I actually find it easier to understand the result using that image than the second. Either diagonal splits the rhombus into two triangles, each of area $$\frac{1}{2}\frac{d_1}{2}d_2=\frac{1}{2}d_1\frac{d_2}{2}.$$ (Is there some aspect-ratio issue with your image? The white rectangle should be a square, but doesn't look like it on my screen.)$\endgroup$
– Will OrrickApr 6 '14 at 2:36

21

$\begingroup$The first rhombus picture is fine. Just draw a rectangle around it, and it's plain to see that the four triangles that form the rhombus cover half of the rectangle. In contrast, the second picture is so confusing that you yourself got it wrong: you write that it "illustrates why the product of the diagonals equals half of the area of the rhombus", while the ratio is the opposite.$\endgroup$
– LaCApr 7 '14 at 3:47

5

$\begingroup$@Peter - I'll admit that I'm using the word derive a bit loosely for a mathematics forum (it's a soft answer to a soft question), but it's too bad you see this as "dumbing down" and "a waste of time." Most 7th-grade teachers charged with teaching volume would simply write the formula on the board, and it would be well-forgotten by the end of summer. Her technique might have been weak insofar as mathematical rigor goes, but the pedagogy was very strong. I assure you, this woman was not one to "dumb down" anything; I remember re-learning concepts in 11th-grade that she taught us in Jr High.$\endgroup$
– J.R.Feb 22 '15 at 9:44

He gave us several weights that were labeled and about 4 weights without labels (let's call them $A, B, C, D$). Then he told us we should find out the weight of the unlabeled weights. $A$ was very easy as there was a weight $E$ with weight($A$) = weight($E$).
I think at least two of them had the same weight and we could only get them into balance with a combination of the labeled weights. The last one was harder. We had to put a labeled weight on the side of the last one to get the weight.

Then he told us how this can be solved on paper without having the weights. So he introduced us to the concept of equations. That was a truly amazing day. Such an important concept explained with a neat way.

$\begingroup$I do the same thing... except with drawings instead of an actual scale and weights. I need to get this set.$\endgroup$
– Guillermo GarzaApr 7 '14 at 0:34

14

$\begingroup$@MeemoGarza although drawings are nice, too, I think that an actual device has some advantages: Pupils rather get excited by a real device than by a drawing; it's easier to accept something that you can touch and play with where you instantly get feedback if what you did was correct; as soon as pupils got familiar with the concept of equations, you can explain the physics behind scales ($a_1 \cdot m_1 = a_2 \cdot m_2$ where $a$ is the length of the part of the scale and $m$ is the mass on that side). That way students can see that math is also about describing the real world in a formal way.$\endgroup$
– Martin ThomaApr 7 '14 at 5:58

2

$\begingroup$I agree completely, but I don't have the apparatus. I'll look into purchasing one before the next semester. Thanks for helping me make up my mind!$\endgroup$
– Guillermo GarzaApr 8 '14 at 0:50

$\begingroup$@MeemoGarza Meanwhile you might be interested in this: walter-fendt.de/ph14d/hebel.htm - it's a site created by a physics teacher I had in school. This website contains lots of Java applets that show things in physics.$\endgroup$
– Martin ThomaApr 8 '14 at 1:59

$\begingroup$This would be amazing if you animated the change from the first to the second by having the triangles on the right flop over onto the ones on the left to create those rectangles.$\endgroup$
– Nick RetallackApr 7 '14 at 0:31

2

$\begingroup$I think this one is lacking because it is not explained in the left picture that the blue and red areas are the same size.$\endgroup$
– SparrApr 7 '14 at 15:58

2

$\begingroup$@Sparr they are not. The salmon area is just 4 times the size of the triangle. The only equivalence is bewteen the blue area on the left (c²) and the blue area on the right (a² + b²) which is just Pythagoras' formula. I will upload a better picture to show it.$\endgroup$
– TobiaApr 7 '14 at 16:33

$\begingroup$Anyway...THIS was my favourite, so, good that you already did it!$\endgroup$
– MattAllegroMay 12 '14 at 21:49

$\begingroup$Awesome — this will be a great introduction to fields & the usage of integrals which I am about to explain to a friend of mine! ;)$\endgroup$
– LukeAug 27 '14 at 17:44

4

$\begingroup$Nice visual intuition, but why is it true that $|\mathbf{r}'|$ is the right integrator to use? (i.e. the 'flattening/projection to the real line' step)$\endgroup$
– enthdegreeAug 11 '16 at 5:18

$\begingroup$Isn't this how students usually learn line integrals?$\endgroup$
– user3932000Mar 9 '17 at 1:50

$\begingroup$I think this one better. In the square one you have to make a choice between dividing horizontally or vertically. With the triangle there is only one dividing cut.$\endgroup$
– abnryApr 5 '14 at 13:55

$\begingroup$Just nitpicking, but you got the areas of the triangles wrong. For example, the one of are $\frac{1}{2}$ according to you has actually an area of $\frac{1}{2}\times\frac{\sqrt{2}}{2}\times\frac{\sqrt{2}}{2}=\frac{1}{4}$, so what this figure is proving is that $\sum_{n=2}^{\infty}\frac{1}{2^n}=\frac{1}{2}$.$\endgroup$
– S4MApr 7 '14 at 7:12

3

$\begingroup$@S4M: The area of the first triangle is defined as $1/2$; how is the calculation wrong when there is no calculation... only definition?$\endgroup$
– Andrew CoonceApr 7 '14 at 16:51

$\begingroup$@S4M: Each new triangle bisects the previous one. The full triangle has area 1. The $\frac{1}{4}$ triangle bisects the triangle of area $\frac{1}{2}$ that is left after the $\frac{1}{2}$ triangle and so on. I'm not trying to prove that the right side of the largest bisection line equals the other side (the summation you have), but that the entire collection of triangles approaches 1.$\endgroup$
– user121055Apr 8 '14 at 21:17

1

$\begingroup$@AndrewCoonce I was wrong in my previous comment. I asscumend that the side of the main triangle was $1$. It's all fine if its side is $\frac{\sqrt{2}}{2}$.$\endgroup$
– S4MApr 9 '14 at 12:27

$\begingroup$But can you show that a sphere with radius, $r$, has a volume of $V_3(r) = \frac{4}{3} \pi r^3$ and a surface area of $SA_3(r) = 4 \pi r^2$$\endgroup$
– Cole JohnsonApr 4 '14 at 16:09

$\begingroup$@ColeJohnson I don't know, just wanted to show this awesome animation:D$\endgroup$
– totymedliApr 4 '14 at 21:11

$\begingroup$I just wish someone could make an animation of that. I understand the math behind it, but animations are nice.$\endgroup$
– Cole JohnsonApr 4 '14 at 22:28

44

$\begingroup$That the circle's perimeter is $2\pi r$ is the definition of $\pi$, so I wouldn't say this is an explanation of the fact; rather it's an illustration of what the definition means, and that the value of $\pi$ is about $3.1$.$\endgroup$
– ShreevatsaRApr 5 '14 at 5:24

$\begingroup$@ColeJohnson There's a neat way to find the volume of a sphere with Cavalieri's principle and Pythagoras. Just look up Cavalieri's principle on Wikipedia, I think it's on there.$\endgroup$
– Akiva WeinbergerFeb 23 '15 at 1:45

$\begingroup$Probably not. But the idea that an object "looks the same if you zoom in on it" is easy to explain, I think.$\endgroup$
– user134824Mar 31 '14 at 17:16

9

$\begingroup$Whereas complex numbers might not be easy to explain, it is easy to explain the idea idea that each pixel is calculated by repeating a simple calculation on two numbers (which are initially the coordinates of the pixel), and the color is a measure of how the resulting pair of numbers "escapes" (grows large).$\endgroup$
– KazApr 1 '14 at 4:57

4

$\begingroup$@adam.r I don't recall precisely when I learned about complex numbers in school, but it was around the 9th grade (maybe 8th, maybe 10th). To understand the Mandelbrot set, nothing more is needed than basic arithmetic with complex numbers. That should put it within reach for anyone who graduated high school. It's certainly a good candidate for this list.$\endgroup$
– SzabolcsApr 1 '14 at 16:12

3

$\begingroup$It is a close up. The Mandelbrot set is not completely self-similar, but there are some self-similarities as you zoom. See this zoom sequence for an example.$\endgroup$
– user134824Apr 3 '14 at 3:12

An iron pendulum is suspended above a flat surface, with three magnets on it. The magnets are colored red, yellow and blue.

We hold the pendulum above a random point of the surface and let it go, holding our finger on the starting point. After some swinging this way and that, under the attractions of the magnets and gravity, it will come to rest over one of the magnets. We color the starting point (under our finger) with the color of the magnet.

Repeating this for every point on the surface, we get the image shown above.

$\begingroup$How is the shading of the image given?$\endgroup$
– David ZhangApr 21 '14 at 14:27

3

$\begingroup$@DavidZhang, the longer the pendulum swings, the darker the point. I don't know the exact function used for this image, but it you click on the link, you'll find the algorithm.$\endgroup$
– PeterApr 21 '14 at 15:27

$\begingroup$Perhaps I'm trying to oversimplify, but this visual proof would be way easier (perhaps even trivial) if the legs of the big straight angle (S.A.) triangle (hypotenuse = the circle's diameter, third vertex on the top of that leg of length $\;b\;$) were drawn, and then from basic geometry: " In a S.A. triangle, the height to the hypotenuse divides the triangle in two triangles similar to each other and also to the big triangle". Nicely brought. +1$\endgroup$
– DonAntonioMar 31 '14 at 12:31

Topology needs to be represented here, specifically knot theory. The following picture is from the Wikipedia page about Seifert Surfaces and was contributed by Accelerometer. Every link (or knot) is the boundary of a smooth orientable surface in 3D-space. This fact is attributed to Herbert Seifert, since he was the first to give an algorithm for constructing them. The surface we are looking at is bounded by Borromean rings.

$\begingroup$I do not know much about topology so I will take your word for it that this is a beautiful idea/concept. However this picture and your description explain nothing to me. It seems like you missed the "easy to explain" bit in the question.$\endgroup$
– dfcApr 7 '14 at 5:25

4

$\begingroup$@dfc I don't know, it seems like you can convey most of the meat here using soap bubbles.$\endgroup$
– SladeApr 7 '14 at 20:10

Fourier transform of the light intensity due to a diffraction pattern caused by light going through 8 pinholes and interfering on a wall, for different choices of parameter:

The best thing about them is, they satisfy periodic boundary conditions, and so you can pick one of them and set it as a desktop background by tiling it, resulting in a far more spectacular image than just the single unit cells posted above!

The images seem to be a vast interconnected network of lines once you tile them, but in fact the entire picture is actually just a single circle, which has been aliased into a tiling cell thousands of times.

Here is the Mathematica code used to generate and save the images. There are two parameters that are adjustable: mag is the magnification and must be an integer, with 1 generating 600 by 600 images, 2 generating 1200 by 1200 images, etc. i is a parameter which can be any real number between 0 and ~1000, with values between 0 and 500 being typical (most of the preceding images used i values between 200 and 300). By varying i, thousands of unique diagrams can be created. Small values of i create simple patterns (low degree of aliasing), and large values generate complex patterns (high degree of aliasing).

$\begingroup$Those images are incredibly beautiful, but can you explain what exactly they represent and how they were generated? "Fourier transform of the light intensity due to a diffraction pattern caused by light going through 8 pinholes and interfering on a wall" isn't very clear for me.$\endgroup$
– gregschlomApr 7 '14 at 9:11

$\begingroup$@Rahul: It's aliasing of a circle. Aliasing is easy to explain. Draw a big circle on a clear plastic sheet. Cut the image into little squares. Stack the squares on top of each other, and look at it. That's the image. The different images above were done using little squares with various side-lengths. I can post the code if you'd like, there are literally tens of thousands of visually distinct diagrams which can be formed.$\endgroup$
– DumpsterDoofusApr 8 '14 at 1:17

$\begingroup$I like this, but one tiny criticism is that it's not visually obvious why the unrolled rings should form a triangle. It's obvious if you can see that the unrolled length of the rings is linearly proportional to their radii, but not visually obvious.$\endgroup$
– LqueryvgSep 1 '14 at 10:52

15

$\begingroup$Is it obvious that you can straighten a ring out and get a rectangle when the ring started out bent? I think a lot of these "bang! pi r squared!" ones leave you with real analysis-induced heebie jeebies :)$\endgroup$
– msouthSep 26 '14 at 0:27

$\begingroup$The rings get smaller as you move in, narrowing to a point. It doesn't matter what the proportion is. It's a simple fact Demonstated by two things. 1. The circles smoothly decrease in length. 2. They narrow to a point. A better way to do this image would be to stack the rings. Then it makes more sense.... Although one can always just rotate by 90 degrees. XD$\endgroup$
– The Great DuckApr 12 '16 at 18:20

$\begingroup$In E-M wave propagation, I thought the E and M components were supposed to be $90$ degrees out of phase. The figure appears to have them in phase or $180$ degrees out of phase, depending on how you interpret the directions of the axes.$\endgroup$
– David KAug 28 '14 at 21:25

1

$\begingroup$@DavidK That's only for circularly polarized photons. This is illustrating a plane-polarized wave, whose components are in phase but offset $90^\circ$.$\endgroup$
– Mario CarneiroJan 9 '15 at 1:42

$\begingroup$@MarioCarneiro Quite right, a quick review of wave equations verify the diagram as shown above. I have this mental image of one of my high school teachers walking across the room waving one arm up and down and one arm side to side to simulate the $E$ and $B$ components, but apparently I don't actually remember it very well after all.$\endgroup$
– David KJan 9 '15 at 2:56

13

$\begingroup$When I read the number 142857 it was like a dejavu for me and I was like "OH MY GOED!! I'VE KNOWN THIS NUMBER!"$\endgroup$
– QwertyJan 11 '15 at 4:20

2

$\begingroup$@NateGlenn It's really just a visualisation of long multiplication. (Try writing out the long multiplication of $13 \cdot 12$ and see how it corresponds to the diagram.)$\endgroup$
– Martin EnderMar 16 '16 at 15:57

I do not know if this meets your criteria of "visually stunning", but nonetheless -

I like this proof of Pythagoras' Theorem (image taken from www.wisfaq.nl):

The key to understanding this is to realize that the inner quadrilateral must be a square - the sides are equal in length (obviously) and each of its angles is $90^{\circ}$ because the two angles on either side sum to $90^{\circ}$, and the sum of the three angles is $180^{\circ}$. The area of this square is $c^2$.

The outer square's area is $(a + b)^2$, which is $c^2$ plus $2 a b$ which is the total area of the four triangles, each of area $\frac{1}{2} a b$.

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